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Vectors
CHAPTER OUTLINE
3.1 Coordinate Systems 3.2 Vector and Scalar Quantities 3.3 Some Properties of Vectors 3.4 Components of a Vector and Unit Vectors
ANSWERS TO QUESTIONS
Q3.1 Only force and velocity are vectors. None of the other quantities requires a direction to be described. The answers are (a) yes (b) no (c) no (d) no (e) no (f ) yes (g) no.
Q3.2 The book’s displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters.
*Q3.3 The vector − 2
D 1 will be twice as long as
D 1 and in the opposite direction, namely northeast. Adding
D 2 , which is about equally long and southwest, we get a sum that is still longer and due east, choice (a).
*Q3.4 The magnitudes of the vectors being added are constant, and we are considering the magnitude only—not the direction—of the resultant. So we need look only at the angle between the vectors being added in each case. The smaller this angle, the larger the resultant magnitude. Thus the ranking is c = e > a > d > b.
*Q3.5 (a) leftward: negative. (b) upward: positive (c) rightward: positive (d) downward: negative (e) Depending on the signs and angles of
A and
B , the sum could be in any quadrant. (f) Now −
A will be in the fourth quadrant, so − +
A B will be in the fourth quadrant.
*Q3.6 (i) The magnitude is 10 2 + 102 m s , answer (f). (ii) Having no y component means answer (a).
*Q3.7 The vertical component is opposite the 30° angle, so sin 30° = (vertical component)/50 m and the answer is (h).
*Q3.8 Take the difference of the coordinates of the ends of the vector. Final first means head end first. (i) − 4 − 2 = −6 cm, answer (j) (ii) 1 − (−2) = 3 cm, answer (c)
Q3.9 (i) If the direction-angle of
A is between 180 degrees and 270 degrees, its components are both negative: answer (c). If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs: answer (b) or (d).
Q3.10 Vectors
A and
B are perpendicular to each other.
Q3.11 No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.
Q3.12 Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds.
45
SOLUTIONS TO PROBLEMS
Section 3.1 Coordinate Systems
P3.1 x = r cos θ = ( 5 50. m ) cos 240 ° = ( 5 50. m ) −( 0 5.) = −2 7. 5 5 m
y = r sin θ = ( 5 50. m ) sin 240 ° = ( 5 50. m ) −( 0 8. 66 ) = − 4 ..76 m
P3.2 (a) x = r cos θ and y = r sin θ , therefore
x 1 = ( 2 50. m )cos 30 0.°, y 1 = ( 2 50. m )sin 30 0.°, and
(^ x^ 1 ,^ y 1 ) =^ (^ 2 17.^ ,^ 1 25.^ )m
x 2 = ( 3 80. m) cos 120 °, y 2 = ( 3 80. m (^) ) sin 120 °, and
(^ x^ 2 ,^ y 2 ) =^ ( −1 90.^ ,^ 3 29.^ )m
(b) d = ( ∆ x ) 2 + ( ∆ y ) 2 = 4 07. 2 + 2 04. 2 m = 4 55. m
P3.3 The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m.
(a) We can use the Pythagorean theorem to find the distance from the origin to the fly.
distance = x^2 + y^2 = ( 2 00. m ) + (^2 1 00. m ) 2 = 5 00. m 2 = 2 2 24. m
(b) θ = ⎛ ⎝⎜^
tan− 1 .°
r = 2 24. m, 26 6.°
P3.4 We have 2 00. = r cos30 0.°
r = °
cos.
and y = r sin 30 0. ° = 2 31. sin 30 0. ° = 1 15..
P3.5 We have r = x^2 + y^2 and θ = ⎛ ⎝⎜^
tan− 1 y x
(a) The radius for this new point is
(^ − x^ ) 2 +^ y^^2 =^ x^^2 +^ y^^2 = r
and its angle is
tan− −
y x
° −θ (^).
(b) ( − 2 x ) 2 + −( 2 y )^2 = 2 r This point is in the third quadrant if (^) ( x y , ) is in the first quadrant or in the fourth quadrant if (^) ( x y , ) is in the second quadrant. It is at an angle of 180° +θ.
(c) ( 3 x ) 2 + −( 3 y )^2 = 3 r This point is in the fourth quadrant if (^) ( x y , ) is in the first quadrant or in the third quadrant if (^) ( x y , ) is in the second quadrant. It is at an angle of −θ.
46 Chapter 3
48 Chapter 3
P3.9 (a)
d = − 10 0. i ˆ^ = 10 0. m since the displacement is in a straight line from point A to point B.
(b) The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle ( ACB ).
s = (^) ( r ) = =
2 π 5 π 15 7. m
(c) If the circle is complete,
d begins and ends at point A. Hence,
d = 0.
P3.10 (a) The large majority of people are standing or sitting at this hour. Their instantaneous foot- to-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 10 5 m upward.
(b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, and west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as ~ 10 5 (0 03. m ) + 10 2 ( 1 m) ~ 10 3 m upward.
P3.11 To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0 5. m)
(a)
A +
B = 5.2 m at 60°
(b)
A −
B = 3.0 m at 330°
(c)
B −
A = 3.0 m at 150°
(d)
A − 2 B = 5.2 m at 300°
C
B A
5.00 m d
FIG. P3.
FIG. P3.
Vectors 49
P3.12 The three diagrams shown below represent the graphical solutions for the three vector sums: R (^) 1 = A + B + C ,
R (^) 2 = B + C + A , and
R (^) 3 = C + B + A. We observe that
R 1 = R 2 = R 3 ,
illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added.
2
B
R (^1) A C
C C
R R 3
A (^) A B
B
100 m
FIG. P3.
P3.13 The scale drawing for the graphical solution should be similar to the fi gure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be
d = 420 ft and θ = − 3 °.
Section 3.4 Components of a Vector and Unit Vectors
*P3.14 We assume the floor is level. Take the x axis in the direction of the first displacement.
If both of the 90° turns are to the right or both to the left , the displacements add like
40 0. m ˆ i^ + 15 0. m ˆ j^ − 20 0. mˆ i^ = (^) ( 20 0. ˆ i^ +15 0. ˆˆ j ) m to give (a) displacement magnitude (20^2 + 152 )^1 ^2 m = 25.0 m at (b) tan−1(1520) = 36.9°.
If one turn is right and the other is left , the displacements add like
40 0. m ˆ i^ + 15 0. m ˆ j^ + 20 0. mˆ i^ = (^) (60 0. ˆ i^ +15 0. ˆˆ j ) m
to give (a) displacement magnitude (60^2 + 152 )^1 ^2 m = 61.8 m
at (b) tan−1(1560) = 14.0˚. Just two answers are possible.
FIG. P3.
Vectors 51
P3.21 Let + x be East and + y be North.
x
y
∑
∑
cos
sin
m
1 12 5
358 12 5 358
(^2 2 2 )
. m
d = ( ∑ x ) + ( ∑ y ) = ( ) + −( .) = m
taan
θ
θ
∑ ∑
y x
d 88 m at 2 00. °S of E
P3.22 The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: d (^) DC east = d (^) DA east + d AC east = 730 cos 5 00. ° − 560 siin 21 0. 527 miles. DC north DA north AC
d = d + d nnorth = 730 sin 5 00. ° + 560 cos21 0. °= 586 miles
By the Pythagorean theorem, d = ( d (^) DC east ) 2 + ( d (^) DC north )^2 = 788 mi.
Then tan θ = = d d
DC north DC east
1 11. and θ = 48 0. °.
Thus, Chicago is 788 miles at 48.0° northeast of Dallas.
P3.23 We have
B = R − A :
A
A
x y
cos120 cm cm
sin RR R
x y
cos. sin..
cm cmm
Therefore, B = [ 115 − −( 75 )] i ˆ^ + [ 80 3. − 130 ] j ˆ^ = (^) ( 190 ˆ i^ −49 7. ˆ j ))
= + =
cm
cm
B 190 49 7 196
2 2
1
tan
θ (^) ⎠⎠ = −14 7. °.
P3.24 (a) See figure to the right.
(b)
C = A + B = 2 00. ˆ i^ + 6 00. j ˆ^ + 3 00. i ˆ^ − 2 00. j ˆ^ = 5 00. i ˆ^ ++ 4 00. ˆ j C = + ⎛ ⎝
25 0. 16 0. at tan −^1 = 6 40.
at 38.7°
D = A − B = 2 00. i ˆ^ + 6 00. ˆ j^ − 3 00. ˆ i^ + 2 00. j ˆ^ = −1 00. ˆ ii + 8 00. j ˆ
D = (^) ( − ) + ( ) −
⎝⎜^
. 2. 2 tan 1. .
at
D = 8 06. at (^) ( 180 ° −82 9. °) = 8 06. at97 2.°
x
FIG. P3.
FIG. P3.
52 Chapter 3
P3.25 (a)
(^ A^ + B^ ) =^ ( 3 i^ ˆ^^ −^2 j ˆ^^ ) + −( i ˆ^^ −^4 ˆ j^ ) =^2 i ˆ^^ −^6 ˆ j
(b)
(^ A^ − B^ ) =^ ( 3 i^ ˆ^^ −^2 j ˆ^^ ) − −( i ˆ^^ −^4 ˆ j^ ) =^4 i ˆ^^ +^2 j ˆ
(c)
A + B = 2 2 + 6 2 = 6 32.
(d)
A − B = 4 2 + 2 2 = 4 47.
(e) θ (^) A + B = −⎛− ⎝
tan 1 = −. =
θ (^) A − B = −⎛ ⎝
tan 1 =.
*P3.26 We take the x axis along the slope uphill. Students, get used to this choice! The y axis is perpen- dicular to the slope, at 35° to the vertical. Then the displacement of the snow makes an angle of 90° − 35° − 20° = 35° with the x axis.
(a) Its component parallel to the surface is 5 m cos 35° = 4.10 m toward the top of the hill.
(b) Its component perpendicular to the surface is 5 m sin 35° = 2.87 m.
P3.
d (^) 1 = (^) (− 3 50. j ˆ^ )m d (^) 2 = 8 20. cos 45 0. ° i ˆ^ + 8 20. sin 45 0. °ˆ j^ = (^) ( 5 80. i ˆ^ + 5. 880 ˆ j ) m d (^) 3 = (^) ( −15 0. ˆ i )^ m R = d (^) 1 + d (^) 2 + d (^) 3 = ( − 15 0. +5 80. (^) ) i ˆ^ + ( 5 80. −3 50. (^) )ˆ j^ == (^) ( − 9 20. ˆ i^ +2 30. ˆ j ) m
(or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is
R = R x^2^ + Ry^2 = ( −9 20. ) + ( 2 2 30. ) 2 = 9 48. m.
The direction is θ = −
arctan =
P3.28 Refer to the sketch
R = A + + C = − − + = −
B i j i i
. ˆ^. ˆ^. ˆ
. ˆ^ .. ˆ
/
1 2
j
R
= ⎡⎣( ) + −( )⎤⎦ = yards
P3.29 East North
x y 0 m 4.00 m 1.41 1. –0.500 –0. +0.914 4. R = x^2 + y^2 at tan −^1 ( y x ) = 4 64. m at 78.6° N ofE
A =^10. 0
B =^ 15 0.
C =^ 50 0.
R
FIG. P3.
P3.36 Let the positive x -direction be eastward, the positive y -direction be vertically upward, and the positive z -direction be southward. The total displacement is then d = (^) ( 4 80. ˆ i^ +4 80. j ˆ^ (^) ) cm + (^) ( 3 70. j ˆ^ −3 70. k ˆ^ ) cm =( 4. 880 ˆ i^ + 8 50. ˆ j^ −3 70. k ˆ) cm.
(a) The magnitude is d = (^) ( 4 80. (^) ) 2 + ( 8 50. (^) ) 2 + −( 3 70. (^) ) 2 cm = 10 4. cm.
(b) Its angle with the y axis follows from cos
θ =
, giving θ = 35 5. °.
P3.37 (a)
A = 8 00. i ˆ^ + 12 0. ˆ j^ −4 00. k ˆ
(b)
B
A
= = i + j − k 4
2 00. ˆ^ 3 00. ˆ^ 1 00. ˆ
(c)
C = − 3 A = − 24 0. i ˆ^ − 36 0. j ˆ^ +12 0. k ˆ
P3.38 The y coordinate of the airplane is constant and equal to 7 60. × 10 3 m whereas the x coordinate is given by x = vi t where vi is the constant speed in the horizontal direction. At t = 30 0. s we have x = 8 04. × 10 3 , so vi = 8 0 40 m 30 s = 268 m s. The position vector as a function of time is P = ( 268 m s (^) ) t i ˆ^ + (^) ( 7 60. × 10 3 m)ˆ j^.
At t = 45 0. s,
P = ⎡⎣1 21. × 10 4 ˆ i^ + 7 60. × 103 ˆ j^ ⎤⎦m. The magnitude is P = (^) ( 1 21 × (^10 4) ) + (^) ( 7 60 × (^10) ) = 1 43 × 10
(^2 3 2 )
.. m. m
and the direction is
θ =
×
×
⎝⎜^
arctan =
3 4 ° above tthe horizontal^.
P3.39 The position vector from radar station to ship is
S = (^) ( 17 3. sin 136 °ˆ i^ +17 3. cos 136 ° j ˆ^ (^) ) km = (^) (12 0. ˆ i − 1 12 4. ˆ j ) km.
From station to plane, the position vector is P = (^) ( 19 6. sin 153 ° i ˆ^ + 19 6. cos 153 °ˆ j^ +2 20. k ˆ)km ,
or P = (^) ( 8 90. i ˆ^ − 17 5. ˆ j^ +2 20. k ˆ^ )km.
(a) To fly to the ship, the plane must undergo displacement
D = S − P = (^) ( 3 12. ˆ i^ + 5 02. j ˆ^ −2 20. k ˆ^ )km.
(b) The distance the plane must travel is
D = = (^) ( ) + ( ) + ( ) =
D 3 12. 2 5 02. 2 2 20. 2 km 6 31. km.
5454 Chapter 3Chapter 3
Vectors 55
55
P3.40 (a)
E = ( 17 0. cm ) cos 27 0. ° i ˆ^ + ( 17 0. cm)sin 27 0. ° j ˆ
E = (^) ( 15 1. i ˆ^ +7 72. ˆ j^ )cm
(b)
F = − ( 17 0. cm ) sin 27 0. ° i ˆ^ + ( 17 0. cm)cos 27 0. °ˆ j
F = (^) (− 7 72. i ˆ^ +15 1. ˆ j )^ cm Note that we do not need to explicitly identify the angle with the positive x axis.
(c)
G = + ( 17 0. cm) sin 27 0. ° i ˆ^ + (17 0. cm)cos 27 0. ° j ˆ
G = (^) ( + 7 72. i ˆ^ +15 1. j ˆ^ )cm
P3.41 Ax = −3 00. , Ay = 2 00.
(a)
A = A (^) x i^ ˆ^ + Ay j ˆ^ = − 3 00. ˆ i^ +2 00. j ˆ
(b)
A = A (^) x^2^ + Ay^2 = ( − ) + ( ) = 2 2 3 00. 2 00. 3 61.
tan
θ = =. (− )
A
A
y x
0 667, tan −^1 (− 0 667. ) = − 33 7.°
θ is in the 2 nd^ quadrant, so θ = 180 ° + ( −33 7. °) = 146 °.
(c) Rx = 0 , Ry = −4 00. ,
R = A + B thus
B = R − A and
B (^) x = R (^) x − Ax = 0 − −( 3 00. ) =3 00. , B (^) y = R (^) y − Ay = − 4 00. − 2 00. = −6 00..
Therefore,
B = 3 00. i ˆ^ −6 00. ˆ j^.
P3.42 The hurricane’s first displacement is
km h
⎛ h ⎝⎜^
) at 60 0. ° N of W, and its second
displacement is
km h
⎛ h ⎝⎜^
) due North. With ˆ i representing east and ˆ j representing
north, its total displacement is:
41 0. cos 60 0. 3 00. ˆ^ 41 0. sin km h
h km h
⎠ (^
) −( i (^) ) +⎛ 6 60 0. ° 3 00. ˆ^ 25 0. 1 50. ˆ ⎝
⎠ (^
⎠ (^
h km h
j h j
6 61 5. km (^) ( −ˆ i^ (^) ) + 144 kmˆ j
with magnitude 61 5 144 157 2 2 (. km ) + ( km ) = km.
P3.43 (a) R
R
x y
. cos.. cos.. . sin
.. sin... . ˆ^.
R i ˆˆ j
(b)
R = ( ) + ( ) =
2 2
1
tan
θ 2 28 7. °
F
y
x
27.0º
G
27.0º
E 27.0º
FIG. P3.
A
y
x
B
45º
C
45º
O
FIG. P3.
Vectors 57
Additional Problems
P3.47 Let θ represent the angle between the directions of
A and
B. Since
A and
^ B^ have the same magnitudes, A ,
B , and
R = A + B form an isosceles triangle in which the angles are 180° −θ ,^ θ 2
, and^ θ 2
The magnitude of
R is then R = A
⎝⎜^
cos
θ . This can be seen from applying the law of cosines to the isosceles triangle and using the fact that B = A. Again,
A , −
B , and
D = A − B form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity
1 2 2
( − ) = 2 ⎛ ⎝⎜^
cos θ sin
θ
gives the magnitude of
D as D = A
⎝⎜^
sin
θ . The problem requires that R = 100 D.
Thus, 2 2
A cos A sin ⎛ θ θ ⎝⎜^
⎝⎜^
⎠⎟^
. This gives tan. θ 2
⎝⎜^
and θ = 1 15. °.
P3.48 Let θ represent the angle between the directions of
A and
B. Since
A and
B have the same magnitudes,
A ,
B , and
R = A + B form an isosceles triangle in which the angles are 180° −θ ,^ θ 2
, and^ θ 2
The magnitude of
R is then R = A
⎝⎜^
cos
θ
. This can
be seen by applying the law of cosines to the isosceles triangle and using the fact that B = A. Again,
A , –
B , and
D = A − B form an isosceles triangle with apex angle θ. Applying the law of cosines and the identity
1 2 2
( − ) = 2 ⎛ ⎝⎜^
cos θ sin
θ
gives the magnitude of
D as D = A
⎝⎜^
sin
θ .
The problem requires that R = nD or cos sin
θ θ 2 2
⎝⎜^
⎝⎜^
n giving θ =
⎝⎜^
2 tan −^11 n
The larger R is to be compared to D , the smaller the angle between
A and
B becomes.
P3.49 The position vector from the ground under the controller of the first airplane is
r 1 = (19 2. km )( cos 25 °) i ˆ^ + (19 2. km)( sin 25 ° )ˆ j ++ ( )
= (^) ( + + )
. ˆ^. ˆ^. ˆ^.
km
km
k
i j k The second is at
r 2 = (17 6. km )( cos 20 °) i ˆ^ + (17 6. km)( sin 20 °) ˆ j ++ ( )
= (^) ( + + )
.1 km
km
. ˆ^. ˆ^. ˆ^.
k
i j k
Now the displacement from the first plane to the second is r 2 (^) − r 1 (^) = (^) (− 0 863. i ˆ^ − 2 09. ˆ j^ +0 3. k ˆ^ )km
with magnitude 0 863 2 09 0 3 2 29 2 2 2 (. ) + (. ) + (. ) =. km.
A
θ R B
A
D
- B
FIG. P3.
FIG. P3.
58 Chapter 3
P3.50 Take the x axis along the tail section of the snake. The displacement from tail to head is
240 m ˆ i^ + ( 420 − 240 ) m cos ( 180 ° − 105 °) ˆ i^ − 180 msinn 75 °ˆ j^ = 287 mˆ i −174 mˆ j.
Its magnitude is 287 174 335 2 2 ( ) + ( ) m = m. From v =
distance ∆ t
, the time for each child’s run is
Inge:
distance m h km s ∆ t = =
( )( )( ) v
12 km m h
s
Olaf:
m s
( )( )( )
∆ t 3 33 m
= 126 s.
Inge wins by 126 − 101 = 25 4. s.
P3.51 Let A represent the distance from island 2 to island 3. The displacement is
A = A at 159°. Represent the displacement from 3 to 1 as
B = B at 298 °. We have 4.76 km at 37 ° + + = 0
A B.
For x components 4 76 37 159 298 0 3 80
. cos cos cos .
km km
° A ° B °
A B
B A
= − km+ For y components 4 76 37 159 298 0 2 86
. sin sin sin .
km km
° A ° B °
0 0 358. A − 0 883. B = 0
(a) We solve by eliminating B by substitution: 2 86 0 358 0 883 8 10 1 99 0 2 86
km + A − (− km+ A ) = kkm km km
A A
A
A 1 17 km
(b) B^ = −8 10^.^ km^ +^ 1 99 7 17.^ (^.^ km^ ) = 6 15. km
P3.52 (a) Rx = 2 00. , Ry = 1 00. , Rz = 3 00.
(b)
R = R (^) x^2^ + R (^) y^2^ + Rz^2 = 4 00. + 1 00. + 9 00. = 14 0. = 3 74.
(c) (^) cos θ x x θ x cos x.
R R
R R
(^1) 57 7° rfrom + x
cos θ y y θ y cos y.
R R
R R
(^1) 74 5° rfrom + y
cos θ z z θ z cos z.
R R
R R
(^1) 36 7° rfrom + z
N
28º B
A
C
69º
37º 1
2
3
E
FIG. P3.
60 Chapter 3
P3.
d i d j d i
1 2
3
cos. ° ˆ ssin. ˆ^ ˆ^. ˆ cos.
j i j d
ˆ i (^) sin. ˆ j (^) ˆ i (^) ˆ j
R d d
1
22 3 4 2
d d i j
R
ˆ ˆ (^) m
−
2
1
m
φ
θ
tan. °
++ φ = 237°
P3.58 d dt
d t dt
r i^ j^ j^ = j
( +^ − ) = + − = − ( )
ˆ (^). m s ˆ jj
The position vector at t = 0 is 4 ˆ i^ + 3 ˆ j. At t = 1 s, the position is 4 ˆ i^ + 1 ˆ j , and so on. The object is moving straight downward at 2 ms, so
d dt
r (^) represents its velocity vector.
P3.59 (a) You start at point A :
r 1 (^) = r (^) A = (^) ( 30 0. i ˆ^ −20 0. ˆ j^ )m. The displacement to B is r (^) B − r A = 60 0. i ˆ^ + 80 0. ˆ j^ − 30 0. i ˆ^ + 20 0. j ˆ^ = 30 0. ˆ i^ + 1100 ˆ j (^).
You cover half of this, (^) (15 0. ˆ i^ +50 0. ˆ j ) to move to r 2 (^) = 30 0. ˆ i^ − 20 0. j ˆ^ + 15 0. i ˆ^ + 50 0. ˆ j^ = 45 0. i ˆ^ +30 0. ˆˆ j. Now the displacement from your current position to C is r C (^) − r 2 (^) = − 10 0. i ˆ^ − 10 0. j ˆ^ − 45 0. i ˆ^ − 30 0. j ˆ^ = −55 0. ˆ ii − 40 0. j ˆ^. You cover one-third, moving to r 3 (^) r (^) 2 r (^) 23 45 0 i 30 0 j i
= + ∆ =. ˆ^ +. ˆ^ + (^) ( −55 0. ˆ^ −40 0. ˆ jj (^) ) = 26 7. i ˆ^ +16 7. ˆ j^. The displacement from where you are to D is r D (^) − r (^) 3 = 40 0. i ˆ^ − 30 0. j ˆ^ − 26 7. i ˆ^ − 16 7. ˆ j^ = 13 3. ˆ i^ − 4 46 7. ˆ j. You traverse one-quarter of it, moving to r 4 (^) r 3 (^) r r (^) 3 i j i
= + ( (^) D − ) =. ˆ^ +. ˆ^ + (^) (13 3. ˆ^ − 4 46 7. ˆ j^ (^) ) = 30 0. ˆ i^ + 5 00. ˆ j.
The displacement from your new location to E is r E (^) − r (^) 4 = − 70 0. i ˆ^ + 60 0. ˆ j^ − 30 0. i ˆ^ − 5 00. ˆ j^ = − 100 ˆ i^ ++ 55 0. ˆ j
of which you cover one-fifth the distance, − 20 0. ˆ i^ +11 0. ˆ j , moving to r 4 (^) + ∆ r (^) 45 = 30 0. ˆ i^ + 5 00. j ˆ^ − 20 0. i ˆ^ + 11 0. ˆ j^ =10 0. ˆ ii + 16 0. ˆ j^.
The treasure is at (10 0^.^ m, 16.0 m)^.
FIG. P3.
continued on next page
Vectors 61
(b) Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to
r r r r r A B A
then to
( r A (^) + r B (^) ) r (^) C r A (^) r B (^) r (^) A r (^) B r C
− ( + )
then to
( r A (^) + r B (^) + r (^) C ) r (^) D r A (^) r (^) B r (^) C r (^) A r B
− ( + + )
r C (^) r D 4
and last to
( r (^) A + r B (^) + r C (^) + r D (^) ) r E (^) r A (^) r (^) B r C (^) r D
− ( + + + ) 4
r (^) A r B (^) r C (^) r D (^) r E 5
This center of mass of the tree distribution is the same location whatever order we take the trees in.
P3.60 (a) Let T represent the force exerted by each child. The x component of the resultant force is
T cos 0 + T cos 120 ° + T cos 240 ° = T ( ) + 1 T ( −0 5. ) + T (−0 5.)) = 0
The y component is
T sin 0 + T sin 120 + T sin 240 = 0 + 0 866. T − 0 866. T = 0. Thus, ∑ =
F 0
(b) If the total force is not zero, it must point in some direction. When each child moves one space clockwise, the whole set of forces acting on the tire turns clockwise by that angle so the total force must turn clockwise by that angle, 360 ° N
. Because each child exerts the same force, the new situation is identical to the old and the net force on the tire must still point in the original direction. But the force cannot have two different directions. The contradiction indicates that we were wrong in supposing that the total force is not zero. The total force must be zero.
FIG. P3.
Vectors 63
P3.14 We assume that the shopping cart stays on the level floor. There are two possibilities. If both of the turns are right or both left, the net displacement is (a) 25.0 m (b) at 36.9°. If one turn is right and one is left, we have (a) 61.8 m (b) at 14.0°.
P3.16 1.31 km north; 2.81 km east
P3.18 (a) 5.00 blocks at 53.1° N of E (b) 13.0 blocks
P3.20 − 25 0. m ˆ i^ +43 3. mˆ j
P3.22 788 mi at 48 0. °north of east
P3.24 (a) see the solution (b) 5.00ˆ i + 4.00ˆ j , 6.40 at 38.7°, –1.00ˆ i + 8.00ˆ j , 8.06 at 97.2°
P3.26 (a) 4.10 m toward the top of the hill (b) 2.87 m
P3.28 42.7 yards
P3.
C = 7 30. cm i ˆ^ −7 20. cm j ˆ
P3.
A + B = (2.60 ˆ i + 4.50 ˆ j )m
P3.34 (a) 2.83 m at θ = 315° (b) 13.4 m at θ = 117°
P3.36 (a) 10.4 cm; (b) 35.5°
P3.38 1 43. × 10 4 m at 32.2° above the horizontal
P3.40 (a) (^) (15 1. ˆ i^ +7 72. ˆ j ) cm (b) (^) ( − 7 72. ˆ i^ +15 1. ˆ j (^) )cm (c) (^) (+ 7 72. ˆ i^ +15 1. ˆ j ) cm
P3.42 157 km
P3.44 (a) a = 5.00 and b = 7.00 (b) For vectors to be equal, all of their components must be equal. A vector equation contains more information than a scalar equation.
P3.46 (a) see the solution (b) 18.3 b (c) 12.4 b at 233° counterclockwise from east
P3.48 2
tan−^ ⎛ ⎝
n ⎠
P3.50 25.4 s
P3.52 (a) 2.00, 1.00, 3.00 (b) 3.74 (c) θ x = 57.7°, θ y = 74.5°, θ z = 36.7°
P3.54 (a) (10 000 − 9 600 cos θ )^1 ^2 cm (b) 270° ; 140 cm (c) 90° ; 20.0 cm (d) They do make sense. The maximum value is attained when
A and
B are in the same direction, and it is 60 cm + 80 cm. The minimum value is attained when
A and
B are in opposite directions, and it is 80 cm − 60 cm.
P3.56 We choose the x axis to the right at 35° above the horizontal and the y axis at 90° counterclock- wise from the x axis. Then each vector has only a single nonzero component. The resultant is 7.87 N at 97.8° counterclockwise from a horizontal line to the right.
P3.58 (^) ( −2 00. m s (^) )ˆ j ; its velocity vector
P3.60 (a) zero (b) see the solution
P3.62 (a)
R (^) 1 = a i ˆ^ + b j ˆ^ ;
R 1 = a^2^ + b^2 (b)
R (^) 2 = a i ˆ^ + b ˆ j^ + c k ˆ^ ;
R 2 = a^2 + b^2 + c^2
